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Rank of the 2-cover space of a knot space [closed]
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It is well known that the order of the first homology group of the double branched cover of a knot $K$ in $S^3$ is the determinant of $K$, i.e. $lvert H_1(B_2(K),mathbbZ) rvert = det (K)$. However, I did not find a formula for its rank. Does anyone know the answer to this question?
algebraic-topology knot-theory
edited Aug 2 at 4:37
Kyle Miller
6,764724
asked Jul 30 at 21:32
Mohamed Ait Nouh
224
closed as unclear what you're asking by Arnaud Mortier, Anubhav Mukherjee, amWhy, max_zorn, Key Flex Jul 31 at 0:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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It is well known that the order of the first homology group of the double branched cover of a knot $K$ in $S^3$ is the determinant of $K$, i.e. $lvert H_1(B_2(K),mathbbZ) rvert = det (K)$. However, I did not find a formula for its rank. Does anyone know the answer to this question?
algebraic-topology knot-theory
edited Aug 2 at 4:37
Kyle Miller
6,764724
asked Jul 30 at 21:32
Mohamed Ait Nouh
224
closed as unclear what you're asking by Arnaud Mortier, Anubhav Mukherjee, amWhy, max_zorn, Key Flex Jul 31 at 0:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
What is " determinant of K"? What is "B_2(k)"?
– Anubhav Mukherjee
Jul 30 at 21:42
No answer for @AnubhavMukherjee ? Here is some possibly relevant material but your question is quite obscure to me: web.northeastern.edu/beasley/MATH7375/Lecture23.pdf
– Arnaud Mortier
Jul 30 at 22:07
1
@AnubhavMukherjee In "An introduction to knot theory" by Lickorish, p. 99, the determinant of a link $L$ is $lvert Delta_L(-1)rvert$, where $Delta_L$ is the Alexander polynomial, or equivalently $lvertdet Grvert$, where $G$ is a Goeritz matrix associated to a double branched cover of $L$. Presumably $B_2(K)$ is a double branched cover based on the context, but that ought to be clarified.
– Kyle Miller
Jul 31 at 2:37
@kyle although I'm familiar with knot theory, but still I didn't get the question after reading those notations. The op should elaborate those. Anyway thank you for the explanation :)
– Anubhav Mukherjee
Jul 31 at 2:41
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up vote
1
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up vote
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down vote
favorite
It is well known that the order of the first homology group of the double branched cover of a knot $K$ in $S^3$ is the determinant of $K$, i.e. $lvert H_1(B_2(K),mathbbZ) rvert = det (K)$. However, I did not find a formula for its rank. Does anyone know the answer to this question?
algebraic-topology knot-theory
edited Aug 2 at 4:37
Kyle Miller
6,764724
asked Jul 30 at 21:32
Mohamed Ait Nouh
224
It is well known that the order of the first homology group of the double branched cover of a knot $K$ in $S^3$ is the determinant of $K$, i.e. $lvert H_1(B_2(K),mathbbZ) rvert = det (K)$. However, I did not find a formula for its rank. Does anyone know the answer to this question?
algebraic-topology knot-theory
edited Aug 2 at 4:37
Kyle Miller
6,764724
asked Jul 30 at 21:32
Mohamed Ait Nouh
224
edited Aug 2 at 4:37
Kyle Miller
6,764724
edited Aug 2 at 4:37
Kyle Miller
6,764724
edited Aug 2 at 4:37
Kyle Miller
6,764724
6,764724
asked Jul 30 at 21:32
Mohamed Ait Nouh
224
asked Jul 30 at 21:32
Mohamed Ait Nouh
224
asked Jul 30 at 21:32
Mohamed Ait Nouh
224
224
closed as unclear what you're asking by Arnaud Mortier, Anubhav Mukherjee, amWhy, max_zorn, Key Flex Jul 31 at 0:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Arnaud Mortier, Anubhav Mukherjee, amWhy, max_zorn, Key Flex Jul 31 at 0:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
What is " determinant of K"? What is "B_2(k)"?
– Anubhav Mukherjee
Jul 30 at 21:42
No answer for @AnubhavMukherjee ? Here is some possibly relevant material but your question is quite obscure to me: web.northeastern.edu/beasley/MATH7375/Lecture23.pdf
– Arnaud Mortier
Jul 30 at 22:07
1
@AnubhavMukherjee In "An introduction to knot theory" by Lickorish, p. 99, the determinant of a link $L$ is $lvert Delta_L(-1)rvert$, where $Delta_L$ is the Alexander polynomial, or equivalently $lvertdet Grvert$, where $G$ is a Goeritz matrix associated to a double branched cover of $L$. Presumably $B_2(K)$ is a double branched cover based on the context, but that ought to be clarified.
– Kyle Miller
Jul 31 at 2:37
@kyle although I'm familiar with knot theory, but still I didn't get the question after reading those notations. The op should elaborate those. Anyway thank you for the explanation :)
– Anubhav Mukherjee
Jul 31 at 2:41
add a comment |Â
3
What is " determinant of K"? What is "B_2(k)"?
– Anubhav Mukherjee
Jul 30 at 21:42
No answer for @AnubhavMukherjee ? Here is some possibly relevant material but your question is quite obscure to me: web.northeastern.edu/beasley/MATH7375/Lecture23.pdf
– Arnaud Mortier
Jul 30 at 22:07
1
@AnubhavMukherjee In "An introduction to knot theory" by Lickorish, p. 99, the determinant of a link $L$ is $lvert Delta_L(-1)rvert$, where $Delta_L$ is the Alexander polynomial, or equivalently $lvertdet Grvert$, where $G$ is a Goeritz matrix associated to a double branched cover of $L$. Presumably $B_2(K)$ is a double branched cover based on the context, but that ought to be clarified.
– Kyle Miller
Jul 31 at 2:37
@kyle although I'm familiar with knot theory, but still I didn't get the question after reading those notations. The op should elaborate those. Anyway thank you for the explanation :)
– Anubhav Mukherjee
Jul 31 at 2:41
3
3
What is " determinant of K"? What is "B_2(k)"?
– Anubhav Mukherjee
Jul 30 at 21:42
What is " determinant of K"? What is "B_2(k)"?
– Anubhav Mukherjee
Jul 30 at 21:42
No answer for @AnubhavMukherjee ? Here is some possibly relevant material but your question is quite obscure to me: web.northeastern.edu/beasley/MATH7375/Lecture23.pdf
– Arnaud Mortier
Jul 30 at 22:07
No answer for @AnubhavMukherjee ? Here is some possibly relevant material but your question is quite obscure to me: web.northeastern.edu/beasley/MATH7375/Lecture23.pdf
– Arnaud Mortier
Jul 30 at 22:07
1
1
@AnubhavMukherjee In "An introduction to knot theory" by Lickorish, p. 99, the determinant of a link $L$ is $lvert Delta_L(-1)rvert$, where $Delta_L$ is the Alexander polynomial, or equivalently $lvertdet Grvert$, where $G$ is a Goeritz matrix associated to a double branched cover of $L$. Presumably $B_2(K)$ is a double branched cover based on the context, but that ought to be clarified.
– Kyle Miller
Jul 31 at 2:37
@AnubhavMukherjee In "An introduction to knot theory" by Lickorish, p. 99, the determinant of a link $L$ is $lvert Delta_L(-1)rvert$, where $Delta_L$ is the Alexander polynomial, or equivalently $lvertdet Grvert$, where $G$ is a Goeritz matrix associated to a double branched cover of $L$. Presumably $B_2(K)$ is a double branched cover based on the context, but that ought to be clarified.
– Kyle Miller
Jul 31 at 2:37
@kyle although I'm familiar with knot theory, but still I didn't get the question after reading those notations. The op should elaborate those. Anyway thank you for the explanation :)
– Anubhav Mukherjee
Jul 31 at 2:41
@kyle although I'm familiar with knot theory, but still I didn't get the question after reading those notations. The op should elaborate those. Anyway thank you for the explanation :)
– Anubhav Mukherjee
Jul 31 at 2:41
add a comment |Â
1 Answer
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The first homology group of the double branched cover (not the double cover) is closely related to the Alexander polynomial. It can be thought of as the quotient of the Alexander module (the first homology group of the infinite cyclic cover) $mathbbZ[t^pm 1]/(Delta_K(t))$ by the relation that $t=-1$, the second root of unity. Here, $Delta_K(t)$ is the Alexander polynomial of $K$.
Thus, $H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1))$. The knot determinant $lvertDelta_K(-1)rvert$ is the order of this group, and in fact it's just a cyclic group. As a $mathbbZ$-module, that means it is rank-$1$ in the case the the determinant is not $1$, and rank-$0$ if it is $1$.
Double covers, on the other hand, are different. For instance, the first homology of the double cover of the complement of the unknot is still $mathbbZ$, which is infinite order. In comparison, the double branched cover of the unknot is $S^3$, whose first homology is trivial.
answered Jul 31 at 0:32
Kyle Miller
6,764724
Thank you Mr. Kyle Miller! I was not aware of the fact that H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1)). However, I am confused here - if the rank is 1 - then the B_2(K) can not be a homology sphere unless det(K)=1. It is known (Milnor proved it) thet B_2(T(p,q)) is the Brieskorn homology sphere $Sigma (p,q,2)$. This contradicts this formula. Reference maths.ed.ac.uk/~v1ranick/papers/milnbries.pdf
– Mohamed Ait Nouh
Aug 2 at 3:16
This puzzle you present took a while to pin down, but if I'm not mistaken, the paper says it is a homology sphere in the case that both $p$ and $q$ are odd. The determinant of $T(2r+1,2s+1)$ is $1$ by my calculations via the Alexander polynomial. (Chapter 9 of Lickorish's book "An Introduction to Knot Theory" discusses the homology of cyclic branched covers, if you want a reference.)
– Kyle Miller
Aug 2 at 4:35
add a comment |Â
1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The first homology group of the double branched cover (not the double cover) is closely related to the Alexander polynomial. It can be thought of as the quotient of the Alexander module (the first homology group of the infinite cyclic cover) $mathbbZ[t^pm 1]/(Delta_K(t))$ by the relation that $t=-1$, the second root of unity. Here, $Delta_K(t)$ is the Alexander polynomial of $K$.
Thus, $H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1))$. The knot determinant $lvertDelta_K(-1)rvert$ is the order of this group, and in fact it's just a cyclic group. As a $mathbbZ$-module, that means it is rank-$1$ in the case the the determinant is not $1$, and rank-$0$ if it is $1$.
Double covers, on the other hand, are different. For instance, the first homology of the double cover of the complement of the unknot is still $mathbbZ$, which is infinite order. In comparison, the double branched cover of the unknot is $S^3$, whose first homology is trivial.
answered Jul 31 at 0:32
Kyle Miller
6,764724
Thank you Mr. Kyle Miller! I was not aware of the fact that H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1)). However, I am confused here - if the rank is 1 - then the B_2(K) can not be a homology sphere unless det(K)=1. It is known (Milnor proved it) thet B_2(T(p,q)) is the Brieskorn homology sphere $Sigma (p,q,2)$. This contradicts this formula. Reference maths.ed.ac.uk/~v1ranick/papers/milnbries.pdf
– Mohamed Ait Nouh
Aug 2 at 3:16
This puzzle you present took a while to pin down, but if I'm not mistaken, the paper says it is a homology sphere in the case that both $p$ and $q$ are odd. The determinant of $T(2r+1,2s+1)$ is $1$ by my calculations via the Alexander polynomial. (Chapter 9 of Lickorish's book "An Introduction to Knot Theory" discusses the homology of cyclic branched covers, if you want a reference.)
– Kyle Miller
Aug 2 at 4:35
add a comment |Â
up vote
0
down vote
The first homology group of the double branched cover (not the double cover) is closely related to the Alexander polynomial. It can be thought of as the quotient of the Alexander module (the first homology group of the infinite cyclic cover) $mathbbZ[t^pm 1]/(Delta_K(t))$ by the relation that $t=-1$, the second root of unity. Here, $Delta_K(t)$ is the Alexander polynomial of $K$.
Thus, $H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1))$. The knot determinant $lvertDelta_K(-1)rvert$ is the order of this group, and in fact it's just a cyclic group. As a $mathbbZ$-module, that means it is rank-$1$ in the case the the determinant is not $1$, and rank-$0$ if it is $1$.
Double covers, on the other hand, are different. For instance, the first homology of the double cover of the complement of the unknot is still $mathbbZ$, which is infinite order. In comparison, the double branched cover of the unknot is $S^3$, whose first homology is trivial.
answered Jul 31 at 0:32
Kyle Miller
6,764724
Thank you Mr. Kyle Miller! I was not aware of the fact that H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1)). However, I am confused here - if the rank is 1 - then the B_2(K) can not be a homology sphere unless det(K)=1. It is known (Milnor proved it) thet B_2(T(p,q)) is the Brieskorn homology sphere $Sigma (p,q,2)$. This contradicts this formula. Reference maths.ed.ac.uk/~v1ranick/papers/milnbries.pdf
– Mohamed Ait Nouh
Aug 2 at 3:16
This puzzle you present took a while to pin down, but if I'm not mistaken, the paper says it is a homology sphere in the case that both $p$ and $q$ are odd. The determinant of $T(2r+1,2s+1)$ is $1$ by my calculations via the Alexander polynomial. (Chapter 9 of Lickorish's book "An Introduction to Knot Theory" discusses the homology of cyclic branched covers, if you want a reference.)
– Kyle Miller
Aug 2 at 4:35
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The first homology group of the double branched cover (not the double cover) is closely related to the Alexander polynomial. It can be thought of as the quotient of the Alexander module (the first homology group of the infinite cyclic cover) $mathbbZ[t^pm 1]/(Delta_K(t))$ by the relation that $t=-1$, the second root of unity. Here, $Delta_K(t)$ is the Alexander polynomial of $K$.
Thus, $H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1))$. The knot determinant $lvertDelta_K(-1)rvert$ is the order of this group, and in fact it's just a cyclic group. As a $mathbbZ$-module, that means it is rank-$1$ in the case the the determinant is not $1$, and rank-$0$ if it is $1$.
Double covers, on the other hand, are different. For instance, the first homology of the double cover of the complement of the unknot is still $mathbbZ$, which is infinite order. In comparison, the double branched cover of the unknot is $S^3$, whose first homology is trivial.
answered Jul 31 at 0:32
Kyle Miller
6,764724
The first homology group of the double branched cover (not the double cover) is closely related to the Alexander polynomial. It can be thought of as the quotient of the Alexander module (the first homology group of the infinite cyclic cover) $mathbbZ[t^pm 1]/(Delta_K(t))$ by the relation that $t=-1$, the second root of unity. Here, $Delta_K(t)$ is the Alexander polynomial of $K$.
Thus, $H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1))$. The knot determinant $lvertDelta_K(-1)rvert$ is the order of this group, and in fact it's just a cyclic group. As a $mathbbZ$-module, that means it is rank-$1$ in the case the the determinant is not $1$, and rank-$0$ if it is $1$.
Double covers, on the other hand, are different. For instance, the first homology of the double cover of the complement of the unknot is still $mathbbZ$, which is infinite order. In comparison, the double branched cover of the unknot is $S^3$, whose first homology is trivial.
answered Jul 31 at 0:32
Kyle Miller
6,764724
edited Aug 2 at 4:15
answered Jul 31 at 0:32
Kyle Miller
6,764724
answered Jul 31 at 0:32
Kyle Miller
6,764724
answered Jul 31 at 0:32
Kyle Miller
6,764724
6,764724
Thank you Mr. Kyle Miller! I was not aware of the fact that H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1)). However, I am confused here - if the rank is 1 - then the B_2(K) can not be a homology sphere unless det(K)=1. It is known (Milnor proved it) thet B_2(T(p,q)) is the Brieskorn homology sphere $Sigma (p,q,2)$. This contradicts this formula. Reference maths.ed.ac.uk/~v1ranick/papers/milnbries.pdf
– Mohamed Ait Nouh
Aug 2 at 3:16
This puzzle you present took a while to pin down, but if I'm not mistaken, the paper says it is a homology sphere in the case that both $p$ and $q$ are odd. The determinant of $T(2r+1,2s+1)$ is $1$ by my calculations via the Alexander polynomial. (Chapter 9 of Lickorish's book "An Introduction to Knot Theory" discusses the homology of cyclic branched covers, if you want a reference.)
– Kyle Miller
Aug 2 at 4:35
add a comment |Â
Thank you Mr. Kyle Miller! I was not aware of the fact that H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1)). However, I am confused here - if the rank is 1 - then the B_2(K) can not be a homology sphere unless det(K)=1. It is known (Milnor proved it) thet B_2(T(p,q)) is the Brieskorn homology sphere $Sigma (p,q,2)$. This contradicts this formula. Reference maths.ed.ac.uk/~v1ranick/papers/milnbries.pdf
– Mohamed Ait Nouh
Aug 2 at 3:16
This puzzle you present took a while to pin down, but if I'm not mistaken, the paper says it is a homology sphere in the case that both $p$ and $q$ are odd. The determinant of $T(2r+1,2s+1)$ is $1$ by my calculations via the Alexander polynomial. (Chapter 9 of Lickorish's book "An Introduction to Knot Theory" discusses the homology of cyclic branched covers, if you want a reference.)
– Kyle Miller
Aug 2 at 4:35
Thank you Mr. Kyle Miller! I was not aware of the fact that H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1)). However, I am confused here - if the rank is 1 - then the B_2(K) can not be a homology sphere unless det(K)=1. It is known (Milnor proved it) thet B_2(T(p,q)) is the Brieskorn homology sphere $Sigma (p,q,2)$. This contradicts this formula. Reference maths.ed.ac.uk/~v1ranick/papers/milnbries.pdf
– Mohamed Ait Nouh
Aug 2 at 3:16
Thank you Mr. Kyle Miller! I was not aware of the fact that H_1(B_2(K),mathbbZ)cong mathbbZ/(Delta_K(-1)). However, I am confused here - if the rank is 1 - then the B_2(K) can not be a homology sphere unless det(K)=1. It is known (Milnor proved it) thet B_2(T(p,q)) is the Brieskorn homology sphere $Sigma (p,q,2)$. This contradicts this formula. Reference maths.ed.ac.uk/~v1ranick/papers/milnbries.pdf
– Mohamed Ait Nouh
Aug 2 at 3:16
This puzzle you present took a while to pin down, but if I'm not mistaken, the paper says it is a homology sphere in the case that both $p$ and $q$ are odd. The determinant of $T(2r+1,2s+1)$ is $1$ by my calculations via the Alexander polynomial. (Chapter 9 of Lickorish's book "An Introduction to Knot Theory" discusses the homology of cyclic branched covers, if you want a reference.)
– Kyle Miller
Aug 2 at 4:35
This puzzle you present took a while to pin down, but if I'm not mistaken, the paper says it is a homology sphere in the case that both $p$ and $q$ are odd. The determinant of $T(2r+1,2s+1)$ is $1$ by my calculations via the Alexander polynomial. (Chapter 9 of Lickorish's book "An Introduction to Knot Theory" discusses the homology of cyclic branched covers, if you want a reference.)
– Kyle Miller
Aug 2 at 4:35
add a comment |Â
3
What is " determinant of K"? What is "B_2(k)"?
– Anubhav Mukherjee
Jul 30 at 21:42
No answer for @AnubhavMukherjee ? Here is some possibly relevant material but your question is quite obscure to me: web.northeastern.edu/beasley/MATH7375/Lecture23.pdf
– Arnaud Mortier
Jul 30 at 22:07
1
@AnubhavMukherjee In "An introduction to knot theory" by Lickorish, p. 99, the determinant of a link $L$ is $lvert Delta_L(-1)rvert$, where $Delta_L$ is the Alexander polynomial, or equivalently $lvertdet Grvert$, where $G$ is a Goeritz matrix associated to a double branched cover of $L$. Presumably $B_2(K)$ is a double branched cover based on the context, but that ought to be clarified.
– Kyle Miller
Jul 31 at 2:37
@kyle although I'm familiar with knot theory, but still I didn't get the question after reading those notations. The op should elaborate those. Anyway thank you for the explanation :)
– Anubhav Mukherjee
Jul 31 at 2:41